Answer
$m(SO_{2})=7.14g$
Work Step by Step
$V(S_{8})=1l=1dm^{3}$
$T=600^{\circ}C=(600+273.15)K=873.15K$
$p=1atm=101.3kPa$
---$m(SO_{2})=?$
$pV=n(S_{8})RT \implies n(S_{8})=\frac{pV}{RT}=\frac{101.3kPa\times 1dm^{3}}{8.314\frac{J}{K\times mol}\times 873.15K}=0.014mol$
The reaction equation of sulfur vapor with pure oxygen is shown below:
$S_{8}+8O_{2}\rightarrow 8SO_{2}$
We can observe that $1mol$ of $S_{8}$ gives $8mol$ of $SO_{2}$. Therefore, we can obtain $8\times 0.014mol = 0.112mol$ of $SO_2$.
$n(SO_{2})=0.112mol \implies m(SO_{2})=M(SO_{2})\times n(SO_{2})=64\frac{g}{mol}\times 0.112mol=7.14g$
